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Downside 1
A model of this drawback seems in Sheldon Ross’s First Course in Likelihood.
An ordinary deck of 52 taking part in playing cards is shuffled. The playing cards are turned up one after the other till an ace seems. Is the following card—the one following the first ace—extra more likely to be the ace of spades or the 2 of golf equipment?
Downside 2
You open a model new deck of playing cards, and the playing cards are organized in growing order by the fits hearts, golf equipment, diamonds, and spades.
The tenth card, for instance, is the ten of hearts. There are 52 playing cards in all. You shuffle the deck completely. What number of playing cards, on common, will nonetheless be in the identical spot as earlier than the deck was shuffled?
As normal, watch the video for an answer.
You received’t imagine the reply, even after you hear it
Or hold studying.
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“All can be properly for those who use your thoughts on your selections, and thoughts solely your selections.” Since 2007, I’ve devoted my life to sharing the enjoyment of sport idea and arithmetic. MindYourDecisions now has over 1,000 free articles with no adverts because of neighborhood help! Assist out and get early entry to posts with a pledge on Patreon.
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Reply To Ace Of Spades Or Two of Golf equipment Downside
(Just about all posts are transcribed rapidly after I make the movies for them–please let me know if there are any typos/errors and I’ll appropriate them, thanks).
Downside 1
At first it appears the 2 of golf equipment ought to be extra possible. If the primary ace is just not the ace of spades, then both card ought to be equally possible. But when the primary ace is the ace of spades, then solely the 2 of golf equipment can seem. So there ought to be extra methods for the 2 of golf equipment. Nonetheless, this instinct is fallacious, and right here is the precise calculation.
Let’s do the ace of spades case first. A deck of 52 playing cards may be organized in 52! methods. Each ordering additionally may be obtained by arranging the 51 different playing cards, after which inserting the ace of spades (52 positions). For every of the 51! orderings, there’s precisely 1 spot to position the ace of spades to comply with the primary ace. So we’ve:
Pr(ace of spades following first ace)
= (quantity methods for fulfillment)/(methods to rearrange all 52 playing cards)
= 51!/52!
= 1/52
However now comes the attention-grabbing half. The very same logic applies for the calculation of the 2 of spades showing simply after the primary ace! There are 51! methods to do that, divided by the 52! methods to rearrange all playing cards. So there’s a 1/52 likelihood of this occasion occurring.
Each occasions have precisely the identical likelihood! It’s simply as possible for the ace of spades to seem after the primary ace as it’s for the 2 of golf equipment.
One solution to verify this calculation is in reality appropriate is to attempt a easy case. We discovered the chance was 1/52 for a deck of 52 playing cards. By the identical logic, the chance ought to be 2!/3! = 1/3 for a 3 card deck.
Suppose 1 = ace of spades, 2 = two of golf equipment, and three = ace of hearts. There are 6! methods to order the deck.
1, 2, 3
(2 follows first ace)
1, 3, 2
(neither follows first ace)
2, 1, 3
(neither follows first ace)
2, 3, 1
(1 follows first ace)
3, 1, 2
(1 follows first ace)
3, 2, 1
(2 follows first ace)
There’s a 2/6 = 1/3 likelihood that the two follows the primary ace and a 1 follows the primary ace. So the deck of three playing cards confirms our calculations for the deck of 52 playing cards.
Downside 2
In case you shuffle a deck of playing cards, what’s the chance the primary card stays within the first place? If the deck is shuffled correctly, then the primary card ought to be equally more likely to be in any of the 52 positions. Subsequently, the chance the cardboard stays in the identical spot is 1/52.
By comparable reasoning, this is identical chance for the second card staying in the identical place, the third card staying in the identical place, and so forth. Every card has a 1/52 likelihood it’s in the identical place.
Now let’s remedy the puzzle.
Let’s outline a Bernoulli random variable Xok to point if the cardboard in place ok stays in the identical place after shuffling. The variable is the same as 1 if the cardboard is in the identical place (chance 1/52) and it is the same as 0 in any other case.
Since every card has a 1/52 likelihood of staying in the identical place, we’ve:
E(Xok) = Pr(similar place)(1) + Pr(totally different place)(0)
E(Xok) = (1/52)(1) + (51/52)(0)
E(Xok) = 1/52
For a randomly shuffled deck, the anticipated variety of playing cards in the identical place is the anticipated worth of the sum that every card stays in the identical place. So we’ve:
E(playing cards in similar place) = E(X1 + X2 + … + X52)
We are able to simplify this expression. We first use linearity of the expectation operator to get:
E(playing cards in similar place) = E(X1) + E(X2) + … + E(X52)
Then we use the truth that every indicator variable has the identical anticipated worth of 1/52.
E(playing cards in similar place) = 1/52 + 1/52 + … + 1/52
As there are 52 phrases, this sum is 1.
E(playing cards in similar place) = 1
So we count on 1 card on common to be in precisely the identical place after shuffling. Counter-intuitively a random shuffle doesn’t imply that each single card goes to a brand new place! It really means we will count on 1 card to be in the identical place.
What’s the reply if the deck has n playing cards? Surprisingly the reply remains to be 1, and it’s the identical for a deck of any dimension, be it simply 1 card and even 1 million playing cards.
In a deck of n playing cards, every card has a 1/n likelihood of being within the appropriate place. Utilizing the identical method as earlier than, we outline a random variable Xok to be 1 if the okth card is in the identical place and 0 in any other case.
Since every card has a 1/n likelihood of staying in the identical place, we’ve:
E(playing cards in similar place) = E(X1 + X2 + … + Xn)
E(playing cards in similar place) = E(X1) + E(X2) + … + E(Xn)
E(playing cards in similar place) = 1/n + 1/n + … + 1/n
E(playing cards in similar place) = n(1/n)
E(playing cards in similar place) = 1
This can be a somewhat stunning end result about random permutations.
Card shuffling numbers from: Su, Francis E., et al. “Making Historical past by Card Shuffling.” Math Enjoyable Info. https://www.math.hmc.edu/funfacts/ffiles/10002.4-6.shtml
Sources
Downside 1
Math152
https://math152.wordpress.com/2008/11/17/ace-of-spades-or-2-of-clubs/
Reddit
https://www.reddit.com/r/HomeworkHelp/feedback/147seb0/high_school_maths_probability_need_help/
Math StackExchange
https://math.stackexchange.com/questions/1380974/probability-of-the-card-following-first-ace-being-ace-of-spades-or-two-of-clubs
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